Download Class Notes Day 31: Intro to Series

Series: Infinite Sums
Series are a way to make sense of certain types of infinitely long sums. We will
need to be able to do this if we are to attain our goal of approximating transcendental functions by using ‘infinite degree’ polynomials. But before we try to add
together an infinite number of polynomials, we first explore what it means to add
an infinite number of numbers.
Here’s the issue: We know how to add two numbers: a1 + a2 . Using associativity
(and parentheses) we can add three numbers
a1 + ( a2 + a3 )
four numbers
a1 + ( a2 + ( a3 + a4 ))
or even n numbers
a1 + ( a2 + ( a3 + ( a4 + (· · · + ( an−1 + an ) . . . )))).
But where would we start (or end) when trying to add an infinite number of
terms? And does the sum add up to a finite number or not? Since all we know
how to do is add a finite number of terms, we will have to use finite addition and
limits to make sense of the process.
Introduction to Series
OK, enough of this finite stuff. What we want to do is add up the terms of an
∞
infinite sequence { an }∞
n=1 . More precisely, given a sequence { an }n=1 , we can form
the infinite sum
∞
a1 + a2 + a3 + · · · + a n + · · · =
∑ ak
k =1
which is called an infinite series or more simply just a series.
Can we do this? Here are several examples.
∞
(a)
∑ k = 1 + 2 + 3 + 4 + · · · = ∞. The sum is clearly not finite; the series
k =1
diverges.
∞
(b)
1
1 1 1
= 1 + + + + · · · . Do these terms add up to a finite sum?
k
2
3 4
k =1
(c)
1
1 1 1
= 1 + + + + · · · . Do these terms add up to a finite sum?
n
2
2 4 8
n =0
∑
∞
∑
∞
(d)
1
1 1 1
= 1 − + − + · · · . Do these terms add up to a finite sum?
n
(−2)
2 4 8
n =0
∑
math 131
infinite series, part i: introduction
DEFINITION 13.1. To find the sum of an infinite series ∑∞
k=1 ak we form the sequence of
partial sums that are often denoted by Sn .
S1 = a 1
S2 = a 1 + a 2
S3 = a 1 + a 2 + a 3
..
.
n
Sn = a1 + a2 + a3 + · · · + a n =
∑ ak
(Sn is called the nth partial sum of the series)
k =1
If the sequence of partial sums {Sn }∞
n=1 has a limit a limit L (converges), we say that the
series converges to L and we write:
∞
n
k =1
k =1
a = lim Sn = L
∑ ak = nlim
→∞ ∑ k
n→∞
or just
∞
∑ ak = L.
k =1
Otherwise the series diverges.
EXAMPLE 13.1. Here’s a simple example. Find the sum of the series
∞
1
1
1
1 1
+··· ,
= + ++ +
k
2
4
8
16
2
k =1
∑
if it exists.
SOLUTION. We first determine each partial sum and then rewrite it in a more conve-
nient form.
1
1
= 1−
2
2
1 1
3
1
S2 = + = = 1 −
2 4
4
4
1 1 1
7
1
S2 = + + = = 1 −
2 4 8
8
8
..
.
S1 =
1
1 1 1
1
+ + +··· n = 1− n
2 4 8
2
2
n
o∞
1
and
So the sequence of partial sums is {Sn }∞
n = 1 = 1 − 2n
Sn =
n =1
lim Sn = lim 1 −
n→∞
n→∞
1
= 1 − 0 = 1,
2n
where we have used Theorem 13.2 to evaluate the limit. In other words,
∞
1
= 1.
k
k =1 2
∑
Pretty cool!
∞
EXAMPLE 13.2. Here’s a another fun example. Find the sum of the series
it exists.
1
if
2+k
k
k =1
∑
SOLUTION. Using partial fractions (check this)
∞
∞
1
∑ k2 + k = ∑
k =1
k =1
1
1
−
k
k+1
=
1−
1
2
+
1 1
−
2 3
+
1 1
−
3 4
+
1 1
−
4 5
+···
2
3
Notice that most of the terms cancel out. The sum collapses and we see that
1
1 1
1 1
1 1
1
1
1
Sn = 1 −
+
−
+
−
+
−
+···+
−
= 1−
.
2
2 3
3 4
4 5
n
n+1
n+1
Such a sum is called a telescoping sum. We are left with only the first and last terms
in the partial sum. This time
lim Sn = lim 1 −
n→∞
n→∞
In other words,
1
= 1 − 0 = 1.
n+1
∞
1
= 1.
2+k
k
k =1
∑
∞
1
1
−
∑ k k + 2 if it
k =1
exists. This time there will be a few more terms that do not cancel. See if you can figure it
out.
YOU TRY IT 13.1. Try this telescoping sum. Find the sum of the series
EXAMPLE 13.3 (Partial Fractions). Here’s another example that uses partial fractions.
∞
Find the sum of the series
4
if it exists.
2 + 3k + 2
k
k =0
∑
SOLUTION. Since the degree of the numerator is smaller than the degree of the de-
nominator a nd since the denominator factors into linear factors, we can write
4
A
B
Ak + 2A + Bk + B
4
=
+
=
.
=
(k + 1)(k + 2)
k+1
k+2
(k + 1)(k + 2)
k2 + 3k + 2
Solving we get:
0 = A + B.
k’s :
(13.1)
and
constants :
4 = 2A + B.
(13.2)
4 = A.
(13.3)
Subtracting (13.1) from (13.2) gives
Putting A = 4 in (13.1) makes B = −4. So we see that
k2
4
4
4
−
.
=
k
+
1
k
+
2
+ 3k + 2
(Check that this is correct!) This means that
∞
∞ 4
4
4
=
−
∑ k2 + 3k + 2 ∑ k + 1 k + 2
k =0
k =0
which is another telescoping series. This time
4 4
4 4
4 4
4
4
4
Sn =
−
+
−
+
−
+···+
−
= 4−
.
1 2
2 3
3 4
n+1
n+2
n+2
So
lim Sn = lim 4 −
n→∞
In other words,
n→∞
4
= 4 − 0 = 4.
n+2
∞
4
= 4.
2 + 3k + 2
k
k =0
∑
Wow!
EXAMPLE 13.4 (Telescoping). Here’s a more complicated example that uses partial
∞
fractions. Find the sum of the series
∑ ln
k =1
( k + 1)
if it exists.
k
math 131
infinite series, part i: introduction
SOLUTION. We can use a log property to rewrite the partial sum as
n
Sn =
∑ ln
k =1
k+1
k
n
=
∑ ln(k + 1) − ln k
k =1
= (ln 2 − ln 1) + (ln 3 − ln 2) + (ln 4 − ln 3) + · · · + [ln(n + 1) − ln n]
= ln n − ln 1.
Therefore
lim Sn = lim ln n − ln 1 = ∞ (diverges)
n→∞
n→∞
and the series
∞
∑ ln
k =1
k+1
k
diverges.
EXAMPLE 13.5 (Partial Fractions). Here’s a more complicated example that uses partial
∞
fractions. Find the sum of the series
∑
k =0
k2
8
if it exists.
+ 4k + 3
SOLUTION. Since the degree of the numerator is smaller than the degree of the denominator a nd since the denominator factors into linear factors, we can write
8
4
A
B
Ak + 3A + Bk + B
=
=
+
=
.
(k + 1)(k + 3)
k+1
k+3
(k + 1)(k + 3)
k2 + 4k + 3
Solving we get:
k’s :
0 = A + B.
(13.4)
and
constants :
8 = 3A + B.
(13.5)
8 = 2A.
(13.6)
Subtracting (13.4) from (13.5) gives
Putting A = 4 in (13.4) makes B = −4. So we see that
8
4
4
=
−
.
k+1
k+3
k2 + 4k + 3
This means that
∞ 4
4
8
=
−
∑
k+1
k+3
k2 + 4k + 3
k =0
k =0
which is another telescoping series.
Sn = 41 − 34 + 42 − 44 + 43 − 54 + 44 − 45 + · · ·
4
4
4
4
4
4
· · · + n−
1 − n +1 + n − n +2 + n +1 − n +3
∞
∑
4
4
Sn = 4 + 2 − n +
2 − n +3 .
So
lim Sn = lim 6 −
n→∞
In other words,
n→∞
4
4
−
= 6.
n+2
n+3
∞
8
= 6.
2 + 4k + 3
k
k =0
∑
YOU TRY IT 13.2 (Partial fractions). Here are two others that are similar to the last example
in that they use partial fractions. See if you can solve them. Find the sums of these series if
they exist.
∞
1
2 + 7k + 12
k
k =0
∞
1
(b) ∑ 2
k
+
4k
+3
k =0
(a)
∑
Answers: (a) 14 ; (b) 23 .
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Geometric Series
Geometric series are among the simpler with which to work. We will see that we
can determine which ones converge and what their limits are fairly easily.
∞
DEFINITION 13.2. A geometric series is a series that has the form
∑ arn , where a is a real
n =0
constant and r is a real number.
YOU TRY IT 13.3. Here are a few examples. Identify a and r in each.
∞
(a)
∑ 6 · 4n
n =0
∞
(b)
1
n
2
n =0
(c)
∑ 2 · 3− n
∑
∞
Answers: (a) 4; (b) 1/2; (c) 1/3.
n =0
∞
Determining the sum of a geometric series
∑ arn is relatively simple. We begin
n =0
by comparing the nth partial sum Sn with rSn . We find:
Sn = a + ar + ar2 + ar3 + · · · + ar n
rSn =
2
n
3
ar + ar + ar + · · · + ar + ar
(13.7)
n −1
(13.8)
So subtracting (13.8) from (13.7) we obtain
Sn − rSn = a − ar n−1
or
(1 − r ) S n = a (1 − r n +1 .
So
Sn =
a (1 − r n +1 )
.
1−r
(13.9)
We know from the Key Limit Theorem 13.2 that


if |r | < 1

0
n
lim r = 1
if r = 1
n→∞



diverges otherwise.
(13.10)
Thus, putting (13.9) and (13.10) together we find

 a
if |r | < 1
a (1 − r n +1 )
lim Sn = lim
= 1−r
n→∞
n→∞
diverges otherwise.
1−r
So we have proved
∞
THEOREM 13.1 (Geometric Series Test). If |r | < 1, then the geometric series
∑ arn converges
n =0
and
∞
∑ arn =
n =0
∞
If |r | ≥ 1, then the geometric series
a
.
1−r
∑ arn diverges.
n =0
math 131
infinite series, part i: introduction
EXAMPLE 13.6. Here are some examples that get progressively more complex.
∞
(a)
Find the sum of the series
n
2
if it exists.
5
n =0
∑
SOLUTION. In this example a = 1 and r = 52 and |r | < 1. So by Theorem 13.1 the
series converges to 1−1 2 = 35 .
5
∞
(b) Find the sum of the series
∑4
n =0
n
6
if it exists.
7
SOLUTION. In this example a = 4 and r = 76 and |r | < 1. So by Theorem 13.1 the
series converges to 1−4 6 =
7
28
1
Find the sum of the series
∑2
= 28.
∞
(c)
n =0
n
3
if it exists.
2
SOLUTION. In this example a = 2 and r = 23 . Since |r | > 1, by Theorem 13.1 the
series diverges.
∞
(d) Find the sum of the series
∑5
−
n =0
1
2
n +2
if it exists.
SOLUTION. Before we can apply the Geometric Series Test, we have to adjust the
power. Notice that we can rewrite the series using the nth power using
∞
∞
1 2
1 n
5
1 n
1 n +2
=
5
−
−
=
−
5
−
∑
∑4 2 .
∑
2
2
2
n =0
n =0
n =0
∞
Now a =
5
4
1−(− 12 )
=
5
4
5
4
3
2
and r = − 12 and |r | < 1. So by Theorem 13.1 the series converges to
= 56 .
SOLUTION. Alternative Method. Another way that we can approach this problem is to write out the first few terms of the series and identify a and r.
∞
1
∑ 5 −2
n =0
n +2
=
5
5
5
5
+
−
+··· .
−
4
8
16
32
|{z} |{z} |{z} |{z}
a
Now a =
5
4
ar
ar2
ar3
and the ratio of a term to the previous one is r = − 21 and |r | < 1. So
by Theorem 13.1 the series converges to
5
4
1−(− 12 )
=
5
4
3
2
= 56 . I find this easier!
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